Let $h: S^4 \to S^4$ be a homeomorphism, and let $f: S^4 \to S^4$ be a continuous map that is homotopic to h. Prove that $h(x)=f(x)$ has a solution.

Question

Here's what I've done so far (correct me if I'm wrong). Showing that $f(x) = h(x)$ has a solution is equivalent to showing $h^{-1}(f(x)) = x$ has a solution since $h$ is a homeomorphism. Since $f$ is homotopic to $h$, by composing with $h^{-1}$, we get that $h^{-1}\circ f$ is homotopic to the identity. This is where I get stuck. Any suggestions?

Answer 1

Let $\varphi_1,\varphi_2\colon \Bbb S^4\to \Bbb S^4$ be two continuous maps such that $\varphi_1(z)\neq \varphi_2(z)$ for all $z\in \Bbb S^4$. Then $\varphi_1$ is homotopic to $-\text{Id}\circ \varphi_2$.

To prove this consider a homotopy $H\colon \Bbb S^4\times [0,1]\to \Bbb S^4$ as $$H(z,t)=\frac{(1-t)\varphi_1(z)-t\varphi_2(z)}{\big|(1-t)\varphi_1(z)-t\varphi_2(z)\big|}\text{ for }(z,t)\in \Bbb S^4\times [0,1].$$ Note that $H$ is well-defined as $\big|(1-t)\varphi_1(z)-t\varphi_2(z)\big|=0$ gives $(1-t)\varphi_1(z)=t\varphi_2(z)$, i.e., $1-t=(1-t)\big|\varphi_1(z)\big|=t\big|\varphi_2(z)\big|=t$ which implies $t=\frac 12$, and so $\varphi_1(z)=\varphi_2(z)$, a contradiction to the hypothesis.

Notice that $H(z,0)=\varphi_1(z)$ and $H(z,1)=-\varphi_2(z)$ for all $z\in \Bbb S^4$. In other words, $H$ is a homotopy from $\varphi_1$ to $-\text{Id}\circ \varphi_2$.

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Since homotopy preserves degree from the previous fact we have $$\deg \varphi_1=\deg(-\text{Id}\circ \varphi_2)=\deg(-\text{Id})\cdot \deg\varphi_2=(-1)^{4+1}\cdot\deg\varphi_2=-\deg\varphi_2.$$ So, $\deg \varphi_1\neq 0$ implies $\deg \varphi_1\neq \deg \varphi_2$ which futher implies $\varphi_1,\varphi_2$ are not homotopic$($Note that any two degree zero maps $\Bbb S^n\to \Bbb S^n$ are homotopic by Hopf Degree theorem$)$. Thanks to @JohnPalmieri for pointing out my error.

To answer the OP's question take $\varphi_1=f, \varphi_2=h$, since $h$ is a homeomorphism we have $\deg h\cdot \deg h^{-1}=\deg (h\circ h^{-1})=\deg \text{Id}=1\implies \deg h\neq 0$. So we are done.

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$\bullet$ The degree of $-\text{Id}\colon \Bbb S^n\to \Bbb S^n$ is $(-1)^{n+1}$, i.e., it induces a map $H_n(\Bbb S^n;\Bbb Z)\cong \Bbb Z\to \Bbb Z\cong H_n(\Bbb S^4;\Bbb Z)$ which is multiplication by $(-1)^{n+1}$. See here

$\bullet$ Clearly, we can do the same procedure for $\Bbb S^{2n}$ for any $n\geq 1$.